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China Hong Kong Mathematical Olympiad

Geometry Difficulty 5.9 AIME, harder Prove it Hong Kong

A triangle $ABC$ is given. A circle $\Gamma$ passes through vertex $A$ and is tangent to side $BC$ at point $P$ . The circle $\Gamma$ intersects sides $AB$ and $AC$ at points $M$ and $N$ , respectively. Prove that (minor) arcs $\widehat{MP}$ and $\widehat{NP}$ are equal if and only if $\Gamma$ is tangent to the circumcircle of $\triangle ABC$ at $A$ .

Solution

If $AB = AC$ , the result is obvious due to symmetry (both statements are equivalent to $P$ being the midpoint of $BC$ ). WLOG assume $AB < AC$ . Let $D$ be the intersection of the line $BC$ and the tangent at $A$ to the circumcircle $(ABC)$ .

If $\overline{MP}$ and $\overline{NP}$ are equal, then $MN$ is parallel to the tangent at $P$ to $\Gamma$ , which is $BC$ . It follows that
$\angle DAM = \angle DAB = \angle ACB = \angle ANM,$
which implies $DA$ is tangent to $\Gamma$ . Therefore, $\Gamma$ is tangent to $(ABC)$ at $A$ .

Conversely, if $\Gamma$ and $(ABC)$ are tangent at $A$ , then $DA$ is tangent to $\Gamma$ . Therefore, we have
$\angle ANM = \angle DAM = \angle DAB = \angle ACB,$
which implies $MN \parallel BC$ . This shows $P$ is the midpoint of $\overline{MN}$ as desired.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.